Problem 36 - Algebra and Trigonometry

Question

Convert the rectangular coordinates to polar coordinates with \(r>0\) and \(0 \leq \theta<2 \pi\). $$ (-\sqrt{6},-\sqrt{2}) $$

Step-by-Step Solution

Answer

The polar coordinates are \((2\sqrt{2}, \frac{7\pi}{6})\).

Step 1: Understanding Rectangular Coordinates

Rectangular coordinates are represented as \((x, y)\). In this problem, \(x = -\sqrt{6}\) and \(y = -\sqrt{2}\). Our goal is to convert these to polar coordinates, \((r, \theta)\), where \(r\) is the radial distance from the origin, and \(\theta\) is the angle measured from the positive x-axis.

Step 2: Calculating the Radial Distance

The radial distance \(r\) in polar coordinates is calculated using \(r = \sqrt{x^2 + y^2}\). Substitute \(x = -\sqrt{6}\) and \(y = -\sqrt{2}\) into the formula: \[ r = \sqrt{(-\sqrt{6})^2 + (-\sqrt{2})^2} = \sqrt{6 + 2} = \sqrt{8} = 2\sqrt{2}. \] So, \(r = 2\sqrt{2}\).

Step 3: Determining the Angle \(\theta\)

The angle \(\theta\) is calculated using the inverse tangent function: \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\). Substitute \(x = -\sqrt{6}\) and \(y = -\sqrt{2}\) into the formula: \[ \theta = \tan^{-1}\left(\frac{-\sqrt{2}}{-\sqrt{6}}\right). \] Simplify the argument: \(\frac{-\sqrt{2}}{-\sqrt{6}} = \frac{\sqrt{2}}{\sqrt{6}} = \frac{1}{\sqrt{3}}\). Since both \(x\) and \(y\) are negative, the point is in the third quadrant, and the angle should be \(\theta = \pi + \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\). Note that \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\), so \(\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\).

Step 4: Stating the Polar Coordinates

The polar coordinates are \((r, \theta) = (2\sqrt{2}, \frac{7\pi}{6})\). Here, \(r = 2\sqrt{2}\) indicates the distance from the origin, and \(\theta = \frac{7\pi}{6}\) is the angle in radians when measured from the positive x-axis counterclockwise.

Key Concepts

Rectangular Coordinates

Have you ever wondered how the position of a point is determined in a plane? In everyday math, we often use rectangular coordinates, also known as Cartesian coordinates. These are represented as \((x, y)\), where \(x\) and \(y\) are the horizontal and vertical distances from a reference point called the origin.
This system allows us to place any point on a grid by moving horizontally by \(x\) units and vertically by \(y\) units.
For instance, in the problem at hand, you have the point \((-\sqrt{6}, -\sqrt{2})\). This means that starting from the origin, \(-\sqrt{6}\) is the distance to move along the x-axis, and \(-\sqrt{2}\) indicates how far to move along the y-axis. When both values are negative, the point is located in the third quadrant of the coordinate plane. Understanding their position is essential before moving to a polar representation.

Radial Distance

In the world of polar coordinates, one of the key concepts is the radial distance \(r\), which is the distance from the origin to the point.
We calculate radial distance using the Pythagorean theorem since \(r\) forms the hypotenuse of a right triangle with sides \(x\) and \(y\). This gives us the formula \(r = \sqrt{x^2 + y^2}\).
For the point \((-\sqrt{6}, -\sqrt{2})\), substitute the given values into the formula:

• Calculate squares: \((-\sqrt{6})^2 = 6\) and \((-\sqrt{2})^2 = 2\)
• Add them up: \(6 + 2 = 8\)
• Take the square root: \(\sqrt{8} = 2\sqrt{2}\)

Thus, the radial distance \(r\) is \(2\sqrt{2}\). This number tells us how far away the point is from the origin.

Inverse Tangent

When calculating angles in polar coordinates, the inverse tangent, also known as the arctangent, is a valuable trigonometric function. It helps us determine the angle \(\theta\) by using the ratio of the opposite side to the adjacent side in a right triangle.
The formula for \(\theta\) is \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\).
In this exercise:

• Substitute \(x = -\sqrt{6}\) and \(y = -\sqrt{2}\).
• We get \(\frac{-\sqrt{2}}{-\sqrt{6}} = \frac{\sqrt{2}}{\sqrt{3}}\). Simplified, this ratio is \(\frac{1}{\sqrt{3}}\).

Finally, we determine the basic angle \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\).
This initial angle doesn’t yet consider the quadrant in which our point is located.

Angle Calculation

Once you have the basic angle from the inverse tangent, it's crucial to find the correct angle \(\theta\) according to the point's position in the coordinate plane.
Our point \((-\sqrt{6}, -\sqrt{2})\) is in the third quadrant associated with angles larger than \(\pi\) (180 degrees) and less than \(3\pi/2\) (270 degrees).
Therefore, to find \(\theta\), we adjust our basic angle by adding \(\pi\) to account for the third quadrant:

• The basic angle from the inverse tangent is \(\frac{\pi}{6}\).
• Adding \(\pi\), we get \(\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\).

Thus, the angle in polar coordinates is \(\frac{7\pi}{6}\). This step ensures the direction of the point from the positive x-axis is accurately measured.